Ah, thank you! L(30)(.6)+(.2)(30)(.8)=16 L=.702, that's the answer I had written down from class. I'll try and figure out the rest with this new solution, thanks!
By "force up" I mean "normal force." That's just what my teacher refers to it as. The force is applied UPWARDS. It's the slope pushing back on the box, which is UP, so meh. And I already calculated that later on, 30sin37=18 N.
I don't see how that applies to the first part. I plugged it in...
This is at the beginning of the problem.
a. I know it's right, no need to explain.
b. Find the velocity of the block when it returns to the bottom of the ramp on the way back down.
I got 3.38 m/s, is this right?
I used the equal Ff = FnU (force of friction = Force up * coefficient of...
A 3kg block is resting at the bottom of a ramp that is at a 37 degree angle with a coefficient of friction of .2. A spring at the bottom of the hill has a spring constant of 5000N/m. The block compresses the spring 8cm (so .08m) and is then released. The block slides up the ramp and then slides...